Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r (as shown in Fig. I). Now, as shown in Fig. II, the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes:

r / ∛2

r / √2

2r / 3

none of the above

Step-by-Step Solution

At equilibrium, the electrostatic repulsive force ( $F_e$ ) is balanced by the component of gravity acting along the tangent of the arc. For small angles $\theta$ , the condition for equilibrium is $\tan \theta = \frac{F_e}{mg}$ .

Geometry: From the figure, $\tan \theta \approx \sin \theta = \frac{r/2}{y} = \frac{r}{2y}$ , where $y$ is the vertical distance from the point of suspension to the line joining the balls.
Force Equation: Substituting Coulomb's law ( $F_e = \frac{kq^2}{r^2}$ ), we get $\frac{kq^2}{r^2 mg} = \frac{r}{2y}$ .
Proportionality: Rearranging the terms, $r^3 \propto y$ .
New Condition: When clamped at half height, the new vertical distance is $y' = y/2$ . Let the new separation be $r'$ .
Calculation: $\frac{(r')^3}{r^3} = \frac{y'}{y} = \frac{1}{2} \implies r' = \frac{r}{\sqrt{2}}$ .

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